Multivariable Calculus Reference

MA2104 Multivariable Calculus Quadric surfaces Partial Derivatives ˆ Cylinder = infinite prism Basic Vectors x2 a2 ˆ Elliptic paraboloid: ˆ Thm 2 [Clairaut’s theorem]: If fxy and fyx are both continuous on disk containing (a, b) then fxy (a, b) = fyx (a, b) y2 b2 + = ˆ Thm 3 [Tangent plane eqn]: Given surface z = f (x, y) with point (a, b): - normal vector: hfx (a, b), fy (a, b), −1i - tangent plane: z = f (a, b) + fx (a, b)(x−a) + fy (a, b)(y−b) z c ˆ Thm 1: kcuk = |c| kuk ˆ Thm 2: (unit vector in direction of a) = a kak ˆ Thm 3 [Dot product properties]: a·b=b·a (da) · b = d (a · b) = a · (db) a · (b + c) = a · b + a · c 0·a=0 2 (a + b) · c = a · c + b · c a · a = kak ˆ Hyperbolic paraboloid: x2 a2 ˆ Ellipsoid: ˆ Thm 4 [Dot product & angle]: a · b = kak kbk cos θ x2 a2 + y2 b2 + z2 c2 − y2 b2 = z c =1 ˆ Thm 5 [Orthogonality]: a ⊥ b ⇐⇒ a · b = 0 ˆ Component (signed scalar): compa b = kbk cos θ = ˆ Projection (vector): proja b = compa b × a kak = i ˆ Cross product: ha1 , a2 , a3 i × hb1 , b2 , b3 i := a1 b1 ha2 b3 − a3 b2 , a3 b1 − a1 b3 , a1 b2 − a2 b1 i a·b kak a·b a·a a j a2 b2 k a3 = b3 x2 a2 ˆ Elliptic cone: + y2 b2 − z2 c2 =0 Hyperboloid : of one sheet x2 a2 + y2 b2 − z2 c2 =1 ˆ Scalar triple product (= signed vol. of parallelepiped): a1 a2 a3 a · (b × c) := b1 b2 b3 = c1 c2 c3 a1 (b2 c3 − b3 c2 ) , a2 (b3 c1 − b1 c3 ) , a3 (b1 c2 − b2 c1 ) ˆ Thm 10 & 11 [Plane]: n · r = n · r0 ⇐⇒ ax + by + cz = ax0 + by0 + cz0 = d ˆ Thm 13 [Derivative properties for vectors]: d 0 0 dt (r (t) + s (t)) = r (t) + s (t) d 0 dt (c r (t)) = c r (t) d 0 0 dt (f (t) r (t)) = f (t) r (t) + f (t) r (t) d 0 0 dt (r (t) · s (t)) = r (t) · s (t) + r (t) · s (t) d 0 0 dt (r (t) × s (t)) = r (t) × s (t) + r (t) × s (t) ˆ Thm 14 [Arc length]: (length from a to b) = ˆ Vector rotation: 90° anticlockwise: 90° clockwise: Rb a ˆ Hyperboloid : of two sheets 2 x a2 + 2 y b2 − 2 z c2 = −1 ˆ Limit: lim f (x, y) = L (x,y)→(a,b) iff for any  > 0 there exists δ >q0 such that 2 2 |f (x, y) − L| <  whenever 0 < (x − a) + (y − b) < δ ˆ Thm 15: To show limit does not exist, take the limit via two different paths that have different limits ˆ Thm 16 & 17 [Limit theorems]: Limits may be taken into addition, subtraction, multiplication, division ˆ Thm 18 [Squeeze theorem]: If |f (x, y) − L| ≤ g(x, y) ∀(x, y) close to (a, b) and lim g(x, y) = 0 0 kr (t)k dt hx, yi → h−y, xi hx, yi → hy, −xi (x,y)→(a,b) then lim f (x, y) = L ˆ Saddle point: (a, b) is a saddle point if fx (a, b) = fy (a, b) = 0 and every neighbourhood at (a, b) contains points (x, y) ∈ D for which f (x, y) < f (a, b) and points (x, y) ∈ D for which f (x, y) > f (a, b) ˆ Local maximum/minimum and critical points cannot be boundary points ˆ Chain rule: ˆ Absolute maximum: f has an absolute max. at (a, b) if ∀(x, y) ∈ D, f (x, y) ≤ f (a, b) For z = f (x, y) and x = x(t), y = y(t): ∂z ∂f ∂x ∂f ∂y = + ∂s ∂x ∂s ∂y ∂s ∂z Fx (x, y, z) =− ∂x Fz (x, y, z) ∂z Fy (x, y, z) =− ∂y Fz (x, y, z) provided Fz (x, y, z) 6= 0 Limits ˆ Local minimum: (a, b) is a local minimum if f (x, y) ≥ f (a, b) for all points (x, y) near (a, b) Differentiation Techniques ˆ Thm 11 [Implicit differentiation]: Given F (x, y, z) = 0 that defines z implicitly as a function of x and y, then: ˆ Thm 7 [Cross prod. & angle]: ka × bk = kak kbk sin θ ˆ Local maximum: (a, b) is a local maximum if f (x, y) ≤ f (a, b) for all points (x, y) near (a, b) ˆ fx & fy are continuous at (a, b) =⇒ f is diff.able at (a, b) ˆ f is differentiable at (a, b) =⇒ f is continuous at (a, b) For z = f (x, y) and x = x(s, t), y = y(s, t): ˆ given f (x, y) : D → R ˆ Critical point: (a, b) is a critical point if fx (a, b) = fy (a, b) = 0 (If point P is a local maximum/minimum then: fx (P ) and fy (P ) both exist =⇒ P is a critical point) dz ∂f dx ∂f dy = + dt ∂x dt ∂y dt ˆ Thm 6: (a × b) ⊥ a and (a × b) ⊥ b ˆ Thm 8 [Cross product properties]: (a + b) × c = a × c + b × c a × b = − (b × a) a × (b + c) = a × b + a × c (da)×b = d (a × b) = a×(db) ˆ Multivariable differentiability: z = f (x, y) is differentiable at (a, b) if 4z = fx (a, b)4x + fy (a, b)4y + 1 4x + 2 4y (with vanishing 1 and 2 ) i.e. zooming in to (a, b) will make surface approximate tangent plane Critical Points, Minimum, Maximum ˆ Quotient rule: g(x) g 0 (x)h(x) − g(x)h0 (x) f (x) = =⇒ f 0 (x) = 2 h(x) [h(x)] ˆ F (x, y, z) = 0 =⇒ normal vector = hFx , Fy , Fz i Gradient Vectors ˆ Thm 13 [Dir. derivatives]: Du f (x, y) = Of (x, y) · u where Of (x, y) := hfx , fy i = gradient vector at (x, y) and u := direction (as unit vector) ˆ Absolute minimum: f has an absolute min. at (a, b) if ∀(x, y) ∈ D, f (x, y) ≥ f (a, b) ˆ Boundary point of R: point (a, b) such that every disk with center (a, b) both contains points in R and not in R ˆ Closed set: Set that contains all its boundary points ˆ Bounded set: Set that is contained in some (finite) disk ˆ Thm 14 [Extreme Value Theorem]: If f (x, y) is continuous on a closed & bounded set D, then the absolute maximum & minimum must exist ˆ To find absolute maximum/minimum of f with domain D: 1) Find the values of f at all critical points in D 2) Find the extreme values of f on the boundary of D 3) Take largest/smallest of the values of Steps 1 & 2 Lagrange Multipliers ˆ Suppose f (x, y) and g(x, y) are differentiable functions such that Og(x, y) 6= 0 on the constraint curve g(x, y) = k. If (x0 , y0 ) is a (local) maximum/minimum of f (x, y) constrained by g(x, y) = k, then Of (x0 , y0 ) = λOg(x0 , y0 ) for some constant λ (the Lagrange multiplier). ˆ Direction of Of (x, y) = steepest upward direction kOf (x, y)k = steepest upward gradient ˆ Thm 1 [Level curve ⊥ Of ]: 0 6= Of (x0 , y0 ) is normal to the level curve f (x, y) = k that contains (x0 , y0 ) (x,y)→(a,b) ˆ Continuity: f is continuous at (a, b) ⇐⇒ lim f (x, y) = f (a, b) (x,y)→(a,b) Surfaces i.e. the limit exists and the f is valid at (a, b) ˆ Level curve of f (x, y) = horizontal trace (for functions in two vars) = 2-D graph of f (x, y) = k for some constant k Contour plot = numerous level curves on the same graph ˆ Thm 20 & 21 [Continuity theorems]: If two functions are continuous (at (a, b)), then their sum, difference, product, quotient, and composition are continuous too (quotient requires denominator 6= 0) ˆ Level surface of f (x, y, z) = 3-D graph of f (x, y, z) = k for some constant k. ˆ All polynomials, trigonometric, exponential, and rational functions are continuous ˆ Thm 2 [Level surface ⊥ OF ]: 0 6= OF (x0 , y0 , z0 ) is normal to the level surface F (x, y, z) = k that contains (x0 , y0 , z0 ) ˆ To find the maximum/minimum points of f (x, y) constrained by g(x, y) = k, we solve  Of (x0 , y0 ) = λOg(x0 , y0 ) g(x0 , y0 ) = k for x0 , y0 , λ. ˆ Jacobian (3D) of transformation (u, v, w) 7→ (x, y, z): Integration Techniques ∂x ˆ Integration by parts: Z Z dv du u dx = u v − v dx dx dx Area & Volume Integrals Z b Z d Z Z a f (x, y) dx dy c c ∂(x,y) ∂(u,v) b f (x, y) dy dx = f (x, y) dA = R d f (x, y) dA = R C a ˆ Line integral for vector field: If curve C is given by r(t) = hx(t), y(t), z(t)i, a ≤ t ≤ b then: Z Z b F(x, y, z) · dr = F(x(t), y(t), z(t)) · r0 (t) dt = b a C ˆ Region types (polar): Type I: D = {(r, θ) : 0 ≤ a ≤ r ≤ b, g1 (r) ≤ θ ≤ g2 (r)} Type II: D = {(r, θ) : α ≤ θ ≤ β, h1 (θ) ≤ r ≤ h2 (θ)} (P dx + Q dy + R dz) = z ! x2 + y 2 + z 2 ˆ Integrating over a spherical wedge: If E = {(ρ, θ, φ) : 0 ≤ a ≤ ρ ≤ b, α ≤ θ ≤ β, c ≤ φ ≤ d} then: ZZZ f (x, y, z) dV = E Z dZ βZ b f (ρ sin φ cos θ, ρ sin φ sin θ, ρ cos φ) ρ2 sin φ dρ dθ dφ α a ˆ Jacobian (2D) of transformation (u, v) 7→ (x, y): ∂x ∂u ∂y ∂u ∂x ∂v ∂y ∂v Z b Z b P x (t)dt + Qy 0 (t)dt + Rz 0 (t)dt 0 a a a Conservative vector fields ˆ A vector field F is conservative on D iff F = Of for some scalar function f on D f is called the potential function of F ˆ To recover f from F = hfx , fy i, do partial integration of fx to get g(x, y) + h(y) [= f (x, y)] (where h(y) is the unknown integration constant), then differentiate g(x, y) + h(y) w.r.t. y and compare with fy to determine h(y) ρ = x2 + y 2 + z 2 θ = atan2(y, x) z φ = cos−1 ρ c b Answer is its negation when r(t) has opposite orientation p p Z C ˆ Spherical coords. ←→ rectangular coords.: x = ρ sin φ cos θ y = ρ sin φ sin θ z = ρ cos φ = cos ˆ If C is a positively oriented (anticlockwise), piecewise smooth, simple closed curve in the plane, and D is the region bounded by C, and F = hP, Qi then:  Z Z ZZ  ∂Q ∂P F · dr = − dA P dx + Q dy = ∂x ∂y C C D (useful when ∂Q ∂x − ∂P ∂y is simpler than P and Q) ∂(x, y) ∂x ∂y ∂x ∂y = = − ∂(u, v) ∂u ∂v ∂v ∂u ZZ ZZ ∂(x, y) f (x, y) dA = f (x(u, v), y(u, v)) du dv ∂(u, v) R S ˆ Test for conservative field (2D): If F = hP, Qi is a vector field in an open (excludes all boundary points) and simply-connected (has no “holes”) region D and both P and Q have continuous first-order partial derivatives on D then: ∂Q ∂P ≡ ⇐⇒ F is conservative on D ∂x ∂y ˆ Test for conservative field (3D): F = hP, Q, Ri (similar requirements as 2D case): ∂P ∂R ∂Q ∂P ∂R ∂Q F is conservative ≡ , ≡ , ≡ ⇐⇒ on D ∂x ∂y ∂y ∂z ∂z ∂x ˆ Fundamental theorem for line integrals: If F is conservative with potential function f , and C is a smooth curve from point A to point B, then: Z Z F · dr = Of · dr = f (B) − f (A) C C =⇒ line integral for conservative field is path-independent ˆ Two paths with different line integrals but same initial and terminal points =⇒ vector field is not conservative  O= ∂ ∂ ∂ , , ∂x ∂y ∂z  Divergence If F = hP, Q, Ri then: ˆ !!! Look out for holes (÷ 0) – use extended Green’s theorem div F = O · F = ∂P ∂Q ∂R + + ∂x ∂y ∂z ˆ !!! When borrowing other question result, check orientation ˆ Reverse application of Green’s theorem: If A is the area of D, then (choose whichever is convenient): Z Z Z 1 A= (x dy − y dx) x dy = − y dx = 2 C C C Parameterize the boundary curve in terms of t (a ≤ t ≤ b)  Z  dy 1 b dx x(t) (e.g. − y(t) dt ) 2 a dt dt Gauss’ theorem If E is a solid region with piecewise smooth boundary surface S with positive (outward) orientation then: ZZ ZZZ F · dS = div F dV S E Curl If F = hP, Q, Ri then: Surface Integrals  ˆ Parametric form of a surface in R3 : r(u, v) = hx(u, v), y(u, v), z(u, v)i , (u, v) ∈ D a Z ˆ Region types (triple integration): Type I: E = {(x, y, z) : (x, y) ∈ D, u1 (x, y) ≤ z ≤ u2 (x, y)} Type II: E = {(x, y, z) : (y, z) ∈ D, u1 (y, z) ≤ x ≤ u2 (y, z)} Type III: E = {(x, y, z) : (x, z) ∈ D, u1 (x, z) ≤ y ≤ u2 (x, z)} −1 Vector Differential Operator Answer is indep. of orientation and parameterization of r(t) f (r cos θ, r sin θ) r dr dθ α ˆ Using = and appropriate f (x, y) may avoid needing to express x, y in terms of u, v ˆ Line integral for scalar field: If curve C is given by r(t) = hx(t), y(t), z(t)i, a ≤ t ≤ b then: Z Z b f (x, y, z) ds = f (x(t), y(t), z(t)) kr0 (t)k dt ˆ Integrating over a polar rectangle: If R = {(r, θ) : 0 ≤ a ≤ r ≤ b, α ≤ θ ≤ β} then: βZ −1 Line Integrals ˆ Polar coords. ←→ rectangular coords.: p x = r cos θ r = x2 + y 2 y = r sin θ θ = atan2(y, x) Z ∂(u,v) ∂(x,y) a ˆ Region types (double integration): Type I: D = {(x, y) : a ≤ x ≤ b, g1 (x) ≤ y ≤ g2 (x)} Type II: D = {(x, y) : c ≤ y ≤ d, h1 (y) ≤ x ≤ h2 (y)} ZZ ∂x ˆ Consider choosing transformation to make bounds constants ˆ Thm 4 [Fubini’s theorem]: If f is continuous on rectangle R = [a, b] × [c, d] then: ZZ ∂x ∂u ∂v ∂w ∂(x, y, z) ∂y ∂y ∂y = ∂u ∂v ∂w ∂(u, v, w) ∂z ∂z ∂z ∂u ∂v ∂w ZZZ f (x, y, z) dV = ZZZ R ∂(x, y, z) f (x(u, v, w), y(u, v, w), z(u, v, w)) du dv dw ∂(u, v, w) S Green’s Theorem ˆ Smooth surface: A surface that is parameterized by r(u, v) where (u, v) ∈ D, such that ru and rv are continuous and ru × rv 6= 0 ∀ (u, v) ∈ D ˆ Thm 6 [Normal vector of parametric surface]: If a smooth surface S has parameterization r(u, v) = hx(u, v), y(u, v), z(u, v)i , (u, v) ∈ D then: ru (a, b) × rv (a, b) is normal to S at (x(a, b), y(a, b), z(a, b)) ˆ Thm 7 [Surface integral for scalar field]: If a smooth surface S has parameterization r(u, v) = hx(u, v), y(u, v), z(u, v)i , (u, v) ∈ D then: ZZ ZZ f (x, y, z) dS = f (x(u, v), y(u, v), z(u, v)) kru ×rv k dA curl F = O × F = ∂R ∂Q ∂P ∂R ∂Q ∂P − , − , − ∂y ∂z ∂z ∂x ∂x ∂y Stokes’ theorem If S is a surface with a boundary curve C (positively oriented w.r.t. S) then: Z ZZ F · dr = curl F · dS C Trigonometric Formulae D F · dS = S F · n̂ dS = S S Positive orientation of boundary curve: If surface S has unit normals pointing towards you, then the positive orientation of boundary curve C goes anti-clockwise Double angle sin 2x = 2 sin x cos x ˆ Thm 7a [Surface integral special case z = g(x, y)]: cos 2x = cos2 x − sin2 x If S is the surface z = g(x, y) where (x, y) ∈ D then: = 2 cos2 x − 1 s   2  2 ZZ ZZ = 1 − 2 sin2 x ∂g ∂g   2 tan x f (x, y, z) dS = f (x, y, g(x, y)) + +1 dA tan 2x = ∂x ∂y S D 1 − tan2 x Triple angle sin 3x = 3 sin x − 4 sin3 x ˆ Orientable surface: two-sided surface Positive orientation: outward from enclosed region cos 3x = 4 cos3 x − 3 cos x Pythagorean ˆ Thm 6 [Surface integral for vector field]: sin2 x + cos2 x = 1 If a smooth surface S has parameterization tan2 x + 1 = sec2 x r(u, v) = hx(u, v), y(u, v), z(u, v)i , (u, v) ∈ D then: cot2 x + 1 = csc2 x ZZ ZZ ZZ S Integrals Z 1 sin2 x dx = (2x − sin 2x) 4 Z 1 2 cos x dx = (2x + sin 2x) 4 Z tan2 x dx = tan x − x Z 1 (cos 3x − 9 cos x) sin3 x dx = 12 Z 1 cos3 x dx = (sin 3x + 9 sin x) 12 Z 1 sin x cos x dx = − cos2 x + C1 2 1 = sin2 x + C2 2 F · (ru × rv ) dA D ˆ Thm 6a [Surface integral special case z = g(x, y)]: If F = hP, Q, Ri, and S is the surface z = g(x, y) where (x, y) ∈ D, then the flux in the upward orientation:  ZZ ZZ  ∂g ∂g F · dS = −P −Q + R dA ∂x ∂y S D  Sum of angles sin(α ± β) = sin α cos β ± cos α sin β cos(α ± β) = cos α cos β ∓ sin α sin β tan α ± tan β tan(α ± β) = 1 ∓ tan α tan β cot α cot β ∓ 1 cot(α ± β) = cot β ± cot α